Sorry I’m just commenting to do it
Given:
The table of values is
x y
-4 2
-3 5
-2 8
-1 11
To find:
The slope of the line that contains these points.
Solution:
From the given table consider any two point.
Let the line passes through the points (-4,2) and (-3,5). So, the equation of the line is




Using distributive property, we get



Therefore, the required equation of line is
.
Answer:
Step-by-step explanation:





Answer:
domain:{-2, infinity) range:(negative infinity , positive infinity)
Step-by-step explanation:
Domain is all x points and range is all y points.
Combine like terms for 30c +3d